By awoo, history, 7 months ago, translation, In English

Hello Codeforces!

On Nov/03/2023 17:35 (Moscow time) Educational Codeforces Round 157 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Artem Ferume Ilikaev, Ruslan AcidWrongGod Kapralov, Alexey ashmelev Shmelev, Alex fcspartakm Frolov, Dmitry DmitryKlenov Klenov, Dmitry dmitryme Mescheryakov, Elena elena Rogacheva and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Please note that the problems of this round partially intersect with the problems of the Southern and Volga Russian Regional Contest. If you took part in that contest, please refrain from participating in the round.

Good luck to all the participants!

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UPD: Editorial is out

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7 months ago, # |
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My comment is on top (yeee) ... By the way thanks to authors for their hard work <3... Hope we'll enjoy the round.

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7 months ago, # |
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Educational round....ok(finger crossed)

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    7 months ago, # ^ |
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    the crossed finger trick do not worked for me (sad wale emoji)...

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7 months ago, # |
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Wow, So exciting to see this round, I hope it will be interesting <3.

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7 months ago, # |
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It is what it is
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7 months ago, # |
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Nothing scares me more than edu round.

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7 months ago, # |
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Why has the frequency of edu rounds decreased?

Not complaining, just curious!

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7 months ago, # |
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Consistent Authors, Hats off

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7 months ago, # |
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The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Artem Ferume Ilikaev, Ruslan AcidWrongGod Kapralov, Alexey I_Remember_Olya_ashmelev Shmelev, Alex fcspartakm Frolov, Dmitry DmitryKlenov Klenov, Dmitry dmitryme Mescheryakov, Elena elena Rogacheva and me.

I think the writer list on the Contest page doesn't reflect this correctly, it's currently just the usual BledDest-Neon-Roms-adedalic-awoo.

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7 months ago, # |
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Educational Rounds are mathforces af. Would skip this one.

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    7 months ago, # ^ |
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    You said the exact same thing last time as well.

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      7 months ago, # ^ |
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      and I don't even understand what he means, if Edu rounds were mathforces I would've farmed the hell out of it

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        7 months ago, # ^ |
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        Right?!

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        7 months ago, # ^ |
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        i think he failed in maths in his high school :}

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        7 months ago, # ^ |
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        Who are you lil bro and how would you farm Edu rounds if they were mathforces?

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          7 months ago, # ^ |
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          If it were mathforces, I would gain rating over number theory tasks every contest, and take advantage of less implementation and more math/observation. That doesn't seem to happen to me anyways if you look at my contest record.

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            7 months ago, # ^ |
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            isnt almost every div2 just observation and constructive stuff tho lol (maybe im just having recency bias)

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7 months ago, # |
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COME ON everyone!!!

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7 months ago, # |
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I hope, inshallah! I will learn something new from this contest.

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7 months ago, # |
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edu rounds always bring me fear

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7 months ago, # |
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Hoping for the final +14 for purple!

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    7 months ago, # ^ |
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    Ah well, C really threw me off. Guess next contest then.

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7 months ago, # |
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If i do B question first and then A question, will I be in the same rank as the person who solved A problem earlier than me but our final time of two question is same?

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    7 months ago, # ^ |
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    Yes .. the rank will be same for this contest.

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    7 months ago, # ^ |
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    it follows the ICPC rules, which count the number of problems solved and the the sum of the time you solved each problem in minutes (of course + penality if you have).

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7 months ago, # |
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GL everyone

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7 months ago, # |
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Damn that Problem C is a good one. (T_T)

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7 months ago, # |
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Another great problem C in educational round.

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    7 months ago, # ^ |
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    I got TLE! Created 5 arrays according to string sizes.
    Accordingly, there will be 4 pairs that produce even sums ((1, 3), (1, 5), (2, 4), (3, 5)).
    Also, count among themselves meaning among array of length 1, 2, 3, 4, and 5.
    Got TLE for this!

    Any leads how to solve it?

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      7 months ago, # ^ |
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      1) We can observe that two strings i,j will be lucky only if their parity is same since we want even length ticket.

      2) So we can solve separately for odd (1,3,5) and even (2,4) since max length is 5 only. Now we just need to see among even and odd, let me consider even case. The combinations possible are 2+2, 2+4, 4+2, 4+4.

      3) Since we want both halves to have same value, for 2+2 and 4+4 case, we will simply use map to store strings with same value and add it to answer. For example if there 3 strings of length 2 with value as 5 (maybe 23,14,32 etc...), then for each we will add 3 since we can concatenate to itself. This case will be same for 1+1,3+3,5+5.

      4) Now for special cases like 2+4, 4+2 we just need to check what is the relation for example in 2+4 case, let ticket be ab+cdef. We want a+b+c = d+e+f, which is same as a+b = d+e+f-c. So we for every 2 length string we will add number of 4 length strings whose d+e+f-c = value of current 2 length string.

      5) You can consider different cases like this for example in 3+5 case we want number of 5 length strings such that their last 4 — first digit = value of 3 length string.

      Hope this helps

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        7 months ago, # ^ |
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        Yeah, thank you for the answer!
        Upvoted!

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        7 months ago, # ^ |
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        If someone needs clean Implementation for the problem

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          7 months ago, # ^ |
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          I tried submitting this, I cannot seem to think what am I missing in this. I tried to think if I recounted anything but cannot seem converge.

          Help me debug, please

          https://mirror.codeforces.com/contest/1895/submission/231216145

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            7 months ago, # ^ |
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            You are using a single frequency array to store all the counts try segregating them based on the size? Also for (int i = 1; i <= 45; ++i) { ans += total[i] * total[i]; } might overflow

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              7 months ago, # ^ |
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              Thanks! Your comment made me read the question once more. I completely misinterpreted. I was going for any suffix sum that matches the prefix + the sum of other entire sting or vice versa.

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          7 months ago, # ^ |
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          Clean solution! Thank you for sharing!
          Upvoted!

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7 months ago, # |
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E's statement is worse. Furthermore, it would be helpful to have an example or explanation provided for the problem

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implementationforces

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D is goated

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How to solve D?

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    7 months ago, # ^ |
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    you can observe that $$$a_{1} ⊕ a_{i} = b_{1} ⊕... b_{i-1} .$$$ Thereafter try to find $$$b_{1}$$$ bitwise

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7 months ago, # |
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How to solve C?

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7 months ago, # |
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Today's D problem is easy if you remember this problem exists

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    7 months ago, # ^ |
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    ig you are talking about the hard version because easy version can be solved using bitmask dp

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How to choose $$$b_0$$$ in D? Any observations, brute force approaches?

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    7 months ago, # ^ |
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    Hint — You have to choose each bit of b0 independently.

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    7 months ago, # ^ |
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    Determine if each bit of b0 should be 1 or 0 by comparing with how many 1s should exist in the final array.

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    7 months ago, # ^ |
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    First initalize it with 0, and caluclate every $$$b_i$$$. Notice now that if you flip a bit in $$$b_0$$$ for ever $$$b_i$$$ that bit flips as well. We can precalculate the number of apperences for each bit in the solution of correct $$$b$$$. Now we can simply compare the occurences of bits in our current solution and the number of bits in the correct solution. If they are the same we do nothing, else we flip.

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      7 months ago, # ^ |
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      Very smart approach! I used a very stupid 01 trie solution...

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        7 months ago, # ^ |
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        me too but i think it's not stupid and easy to think and implement(

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        7 months ago, # ^ |
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        Could you please explain how does a trie help here?

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          7 months ago, # ^ |
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          Let:

          $$$c_{1} = a_{1}$$$

          $$$c_{2} = a_{1} \oplus a_{2}$$$

          ....

          $$$c_{n - 1} = a_{1} \oplus a_{2} ... \oplus a_{n - 1}$$$

          The trivial solution would be we iterate $$$b_{1}$$$ from $$$0$$$ to $$$n - 1$$$ and find out the maximum value of $$$b_{1} \oplus c_{i} (1 <= i <= n - 1)$$$ by brute force. If this value is less or equal than $$$n - 1$$$, then we successfully find out a valid $$$b$$$.

          However, the algorithm above is $$$O(N^2)$$$ To optimize it, we can use the 0-1 trie. Insert all the $$$c_{i}$$$ to a trie from the highest bit to the lowest bit. Then, we can obtain maximum value of $$$b_{1} \oplus c_{i}$$$ by traversing this trie rather than brute force. The time complexity becomes $$$O(NlogN)$$$

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    7 months ago, # ^ |
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    From b[i]^b[i+1] = a[i], we can deduce b[i] = b[1]^a[1]^a[2]^...a[i-1]. Since, the answer always exists all prefixes a[1], a[1]^a[2], ..., a[1]^a[2]^a[3]^...a[n-1], will be unique. So just fix b[1] and check for maximum and minimum XOR

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    7 months ago, # ^ |
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    Most likely people will tell you a solution which considers each bit separately (it is easier to code), but when this problem was proposed, I solved it in a different way

    Spoiler
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      7 months ago, # ^ |
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      I mean, you can just bruteforce for all values of $$$b_1$$$ in $$$[0,n)$$$ and output the one where $$$max=n-1$$$ and $$$min=0$$$, right? That should be trivial using a binary trie storing all prefix XORs.

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        7 months ago, # ^ |
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        Also i don't think it is required to check the min , checking max is suff ig

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        7 months ago, # ^ |
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        Isn't that the exact same thing as what the parent comment from BledDest says?

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          7 months ago, # ^ |
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          BledDest's method focuses on which index to change to $$$0$$$, my method focuses on the literal value of $$$b_1$$$. Kinda similar I guess.

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            7 months ago, # ^ |
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            BledDest's method focuses on which index to change to 0 , my method focuses on the literal value of b1 . Kinda similar I guess.

            Are we reading the same thing? Quoting him...

            Iterate on the first element in the resulting array (let it be x ).

            The first element of the resulting array IS b1. So he's saying the EXACT same thing. (I hope I am not wrong here, it would indicate my final brain calls have finally passed away)

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              7 months ago, # ^ |
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              For what can come as the first element he is using whatever is in the array, I am using literally the interval $$$[0,n)$$$. Little details do matter

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                7 months ago, # ^ |
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                For what can come as the first element he is using whatever is in the array,

                That is not specified anywhere in his comment?

                I (reasonably) assume he means iterating over x in [0, n) only.

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                  7 months ago, # ^ |
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                  "find the maximum value of $$$x⊕c_i$$$ with descent on the trie"

                  That's the part implying he is using the prefix XOR values as $$$b_1$$$. Yeah I apologise, the difference in detail is too minor (just a small implementation advantage)

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                  7 months ago, # ^ |
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                  Oh no need to apologise, I simply wanted to confirm my understanding, I often second guess myself otherwise.

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      7 months ago, # ^ |
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      How does solving each bit independently gives correct answer? We also need to ensure that those bits need to appear in correct combination with each other. For example,(in bits), 000, 001, 010,011,100. And 000, 001, 010, 000, 111. Both have same frequency of each bit but they occur at different positions.

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        7 months ago, # ^ |
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        If some bit $$$x$$$ has the same frequency of $$$0$$$'s and $$$1$$$'s in the resulting sequence no matter how you choose it in the starting element, then it means that $$$n$$$ is divisible by $$$2^{x+1}$$$; and if we flip it, every number in the sequence just transforms from $$$b_i$$$ to $$$b_i \oplus 2^{x}$$$, which is a bijection when $$$n$$$ is divisible by $$$2^{x+1}$$$.

        So, any such bits in the answer can be flipped without breaking anything.

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          7 months ago, # ^ |
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          My question is different. You are answering when frequency of a bit x is same in a set of numbers. My question is: There can exist 2 sets of numbers where frequency of each bit in the two sets are same but the sets are not same. S1 = {00,01,11}, S2 = {10, 01, 01}.

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            7 months ago, # ^ |
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            Since b[0] ^ b[i] = a[0] ^ a[1] ^ ... ^ a[i — 1], if you fix ith bit for b[0], then ith bit of b[i] is uniquely determined.

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              7 months ago, # ^ |
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              Yes, it is uniquely determined but how will you ensure that it is 0,1,2,3,4,...,n-1 only and not something else?

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                7 months ago, # ^ |
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                Suppose at ith bit there are a $$$ 0's$$$ and b $$$1's$$$ in $$$0,1,...n-1$$$ and p be the number of $$$1's$$$ at ith bit in prefix array. Then you can check if $$$ith$$$ bit of $$$b_{1}$$$ can be 0 if $$$p=b$$$ and $$$a>0$$$ otherwise it will be 1.

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                7 months ago, # ^ |
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                If we know the total quantity of $$$1$$$'s in each bit, we know the sum of all numbers. And for every set that is different from $$${0, 1, 2, \dots, n-1}$$$, its sum is strictly greater than required.

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        7 months ago, # ^ |
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        That's how I reason about it:

        For $$$i$$$-th bit of $$$b_1$$$, the necessary condition is that $$$b_1, b_1 \oplus b_2, b_1 \oplus b_3, ..., b_1 \oplus b_n$$$ has the same number of set bits as sequence $$$0, 1, ..., n - 1$$$.

        The only way for this condition to not be sufficient is if setting any bit for $$$b_1$$$ works.

        That can only happen when there are initially more 1 bits in sequence than 0 bits (specifically there is one more 1 bit), which can't happen due to the nature of $$$0, 1, ..., n - 1$$$ sequence (at any n, number of 1 bits is not greater than number of 0 bits)

        EDIT: i wrote some bullshit. Setting any $$$i$$$-th bit in $$$b_1$$$ works, if n is divisible by the $$$2^{i + 1}$$$, and it doesn't affect the answer, because there's always pair $$$(x, y)$$$ in $$$0, 1, ..., n - 1$$$, such that $$$x \oplus y = 2^{i}$$$

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      7 months ago, # ^ |
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      Considering each bit separately is also much easier to reason about, because the condition "every prefix xor of $$$a$$$ should have same number of set bits as sequence $$$0, 1, ..., n - 1$$$" is sufficient and quite natural to get across

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      7 months ago, # ^ |
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      how we ensure that the elements of the array will be unique?

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        7 months ago, # ^ |
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        for a solution $$$x_0$$$, if the resulting array has any $$$i,j$$$ such that $$$(x_0 \oplus c_i)=(x_0 \oplus c_j)$$$, then $$$c_i = c_j$$$, which means for any value of $$$x$$$, $$$(x \oplus c_i)=(x \oplus c_j)$$$, therefore there won't be any solution at all, which contradicts the constraint in the problem.

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      7 months ago, # ^ |
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      I was very close to what you was thinking but instead of thinking to store in binary tree I think that this may be done by greedily

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Why should problems like C appear in CF?

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    7 months ago, # ^ |
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    is it ad-hoc ? just asking ?

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    7 months ago, # ^ |
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    Can you tell what the approach was for C. I was doing a 0(n2) but it was giving a TLE. Without checking all the combinations how is it possible?

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      7 months ago, # ^ |
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      any target number would be at max 10, iterate over each number

      Ex, 93746

      9 3746 => (3+7+4+6) — (9), here we will check if we have any three-digit sum with this sum

      Similarly, we will check for: 93 746 937 46 9374 6

      Also, we can add a number from the backside as well, so check for suffix as well.

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        7 months ago, # ^ |
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        Used same approach but got wrong ans

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        7 months ago, # ^ |
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        Yes but how will you check in less than O(n^2) that"s my doubt

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    7 months ago, # ^ |
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    Used trie, looking for solution that doesn't use it

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      7 months ago, # ^ |
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      iterate over lengths of parts and then over largest of them => you can deduce sum of digits of smallest one (precalc them all)

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      7 months ago, # ^ |
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      since the maximum number of place is 5 you can just hash all option and then check for the correct ones.

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    7 months ago, # ^ |
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    They should have given a smaller test case to fail the solution where (i,j) and (j,i) both does not satisfy. Wasted too much time thinking it was symmetric.

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7 months ago, # |
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good contest

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https://mirror.codeforces.com/contest/1895/submission/231204890 , Can anyone tell me why I am getting TLE , Sorry I am just dumb !

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7 months ago, # |
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I like problem C, even though I have not solved it because I had overcomplicated implementation.

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7 months ago, # |
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Can anyone hack this? I did a randomized solution to D:

https://mirror.codeforces.com/contest/1895/submission/231206864

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    7 months ago, # ^ |
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    How do you hack a randomized solution? I can get your code to fail against some tests on my local machine, but if I try to hack you then it'll fail since the seed is different on the grading server. Is there any way to determine what the seed would be on the grading server? If not, then I don't think your solution is hack-able.

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      7 months ago, # ^ |
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      If everyone creates an unique test case, then he will be hacked for sure.

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    7 months ago, # ^ |
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    What does randomized solution mean?

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    7 months ago, # ^ |
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    Done

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    7 months ago, # ^ |
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    N=2**16+1 input b[0] = 2 ** 16 b[i]= i xor b[0]

    answer must be: a[i] = i

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Cannot find a good solution for D and bashed trie.

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I'm not sure if this problem https://leetcode.com/problems/decode-xored-permutation/ is like today's problem D. These two questions are too similar. I think some people may have just modified the code a bit and moved it over, but I believe everyone wouldn't do that.

But this really affects my mood. In fact, I was looking for the problem for almost half an hour and I finally found it when the contest has finished for 30 seconds. Hope I will see every problem next round new, not existing.

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    7 months ago, # ^ |
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    I'm going to return to Specialist!!! Ah I'll never take a part in any edu. round anymore!!!!

    screaming

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    7 months ago, # ^ |
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    it's kinda easy when n is odd (you can simply calculate first element knowing xor of all of them), but could not find proper solution when n is even >_<

    was sure that optimized n^2 goes through, but spend toooooo much time making up formulas

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Is there any way to actually calculate b[0] in problem D ? I tried to compute XORs from 0 throught n-1, and it is equal to XORs of all prefixes ^(n times) b[0] . But sadly it doesn't help when n is even.

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    7 months ago, # ^ |
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    I don't exactly know why your solution is correct for odd $$$n$$$, the way I found $$$b[0]$$$ is first to find the prefix XOR of the array $$$a$$$. Then count the number of occurrences of each bit in the prefix XOR array. Compare it with the number of occurrences of each bit from $$$0$$$ to $$$n - 1$$$.

    The bits whose number of occurrences is not equal should be in $$$b[0]$$$. (We can see that if the number of occurrences of some $$$i$$$-th bit is $$$k$$$, then choosing that bit in $$$b[0]$$$ will change its number of occurrences to $$$n - k$$$.)

    The remaining numbers can be calculated easily.

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The frustration of not being able to solve more than 2 tasks in div2 rounds is immense!

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for G, splay tree: TLE Red black tree: AC(500ms)

don't know why :(

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7 months ago, # |
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Two solutions that look similar 231191280 231185745

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How can I solve F?

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    7 months ago, # ^ |
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    Count number of arrays with minimum element $$$<x+k$$$ then subtract out number of arrays with max element $$$<x$$$. To do first part consider the number of difference arrays on $$$n$$$ elements, $$$(2k+1)^{n-1}$$$ then consider that you can shift the array up/down to decide where the minimum element will be with $$$x+k$$$ options. To find what we need to subtract out I did matrix expo.

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    7 months ago, # ^ |
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    Let an array $$$a$$$ be good if

    • $$$a_i \ge 0$$$ and
    • $$$|a_i - a_{i-1}| \le k$$$

    Now the answer is equal to (the number of good arrays that contain an integer in range $$$[0, x+k)$$$) minus (the number of good arrays that only contain integers in range $$$[0, x)$$$).

    The number of good arrays that contain an integer in range $$$[0, x+k)$$$ is equal to $$$(x + k) \cdot (2k + 1)^{n-1}$$$

    Proof:

    The number of good arrays that only contain integers in range $$$[0, x)$$$ can be calculated with fast matrix exponentiation:

    Let $$$A[n]$$$ be an $$$x \times 1$$$ matrix where $$$A[n]_i$$$ is equal to the number of good arrays of length $$$n$$$ that only contain integers in range $$$[0, x)$$$ with final element equal to $$$i$$$.

    The number we want to calculate is $$$A[n]_0 + A[n]_1 + \dots + A[n]_{x-1}$$$.

    Clearly, $$$A[1]_0 = A[1]_1 = \dots = A[1]_{x-1} = 1$$$.

    Let $$$M$$$ an $$$x \times x$$$ matrix: $$$M_{ij} = 1$$$ exactly when $$$|i - j| \le k$$$.

    We can notice that $$$A[i+1] = MA[i]$$$. This means that $$$A[n] = M^{n-1}A[1]$$$.

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      7 months ago, # ^ |
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      Thnak u so much for your detailed explanation!

      But how can you come up with that? I just cannot think of this bijection myself.

      Is there some basic logics under it or any problems similar to this problem?

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      7 months ago, # ^ |
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      I have another way to argue why the number of good arrays contain a element in $$$[0, x + k)$$$ is $$$(x + k)(2k + 1)^{n - 1}$$$.

      consider fixing the difference array of $$$a$$$, clearly there are $$$(2k + 1)^{n - 1}$$$ ways to do this, then for each difference array, we can add the whole array with any integer(that is, translate the array horizontally), and let focus on the minimum of this array, clearly it must be no less than $$$0$$$ and less than $$$x + k$$$ to be a good array, so for a difference array, there are $$$(x + k)$$$ ways to do that, thus the number of good array is $$$(x + k)(2k + 1)^{n - 1}$$$.

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7 months ago, # |
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[ Deleted ]

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7 months ago, # |
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Is there a case for D where the elements in the answer array can satisfy all the XOR properties, and whose sum is equal to the sum of [0, n-1], but is not correct? As in some elements might have duplicates and the max element can exceed n-1. My solution goes over the 2^19 ways to either set the ith bit in the first element on and off for each of the 19 bits. Then I just check the total sum in o(1) time for all of these ways to see if it matches the intended sum.

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    7 months ago, # ^ |
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    Such a counterexample is not possible. The fact that a solution exists guarantees that all arrays you can generate consistent with A have distinct elements. There is only 1 possible sequence of N distinct integers that has the same sum as 0+1+2+..+N-1, so your solution works.

    You can prove this more formally by noticing that given A, the generated sequence B is determined entirely by the choice of the initial element B[0]. More formally, we can define B(n) as the candidate sequence B that starts with n:

    B(n)[0] = n
    B(n)[i] = B[i - 1] ^ A[i - 1] = n ^ A[0] ^ A[1] ^ .. ^ A[i -1]  (1 ≤ i < N)
    

    Then it's clear that B(x)[i] = B(0)[i] ^ x, and therefore B(y)[i] = B(x)[i] ^ (x ^ y).

    It follows that if there is any value of x such all elements of B(x) are distinct (which the problem guarantees), then for all x, all values of B(x) must be distinct, because you can't make two distinct value equal by XOR'ing them with the same constant.

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Why did the problemsetters use setTestCase in the validators for single-test problems C and D (and also in some other problem recently)? Samples look ugly, and the use of this function here is pointless.

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    7 months ago, # ^ |
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    There are no setTestCase calls in either of them. Idk why it became ugly on cf recently, but I also noticed that.

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7 months ago, # |
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Ideas
A There can be two cases 1. When x >= y In this case we can just move along the path and pick the key until we reach the chest. 2. When x < y In this case we can greedily pick the chest when we see it until k + x <= y Then we can traverse the leftover distance twice to reach the chest again. So basically min(k + x, y) + 2 * (y - min(k + x, y))
B To get the min path just sort the array choose first N points as X coordinates of each pair Choose the next N points as Y coordinates.

C A thing to notice is that O + O = E & E + E = E. Keeping this in mind the only way we can get even lengths are := 1 + 1, 1 + 3, 3 + 1, 1 + 5, 5 + 1, 2 + 2, 2 + 4, 4 + 2 To find these pairs fast you can notice for the S(s(1) + s(3)) the sum S[0,1] is nothing but s(1) + s(3)(0) and for S(s(3) + s(1)) the second sum is s(3)(2) + s(1). Similarly form these formulas for all other pairs and store each unique sum is 5 seperate maps (Since size of string is atmost 5).

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awoo For Question D of today's contest, my submission had wrong answer on test 2, Test 2 was part of the samples, but still I got penalty for submitting a code that gave a wrong answer on samples, please look into it

Thank you

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7 months ago, # |
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Took too much time to solve C :(

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7 months ago, # |
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In Problem E, is the card returned to the hand of the player who beat the opponent's card?

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    7 months ago, # ^ |
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    From the statement:

    After a card is beaten, it returns to the hand of the player who played it. It implies that each player always has the same set of cards to play as at the start of the game.

    Does this not answer your question?

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7 months ago, # |
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Please could you explain the G statement? How should I count inversions in the string, from which blues were removed? How many inversions do I have in this string: '0010000'. When should I choose 'blue' cost of a symbol instead of 'red'? Too hard for me to get it from statement

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    7 months ago, # ^ |
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    I would solve D i believe if I would not spend 50 minutes on G and somehow managed to miss the statement. Good job!

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7 months ago, # |
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I've been reading many C solutions but didnt get them :c i had other approach but dindt get ac in contest T~T

Here is my code for C, i think it's more intuitive than others. or maybe not I match every 5-3, 5-1, 3-1, 4-2 lengths. Also count math with itself

Code for C T~T
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7 months ago, # |
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Can anyone please explain C?

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    7 months ago, # ^ |
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    Each combined ticket consists of two snippets ("snippets" meaning strings in the input). There are three possible cases:

    1. The two snippets are equal length, e.g. "123" + "222" = "123222" ($$$1+2+3 = 6 = 2+2+2$$$).
    2. The left snippet is longer than the right, e.g. "1234" + "11" = "123411" ($$$1+2+3=6=4+1+1$$$).
    3. The right snippet is longer than the left, e.g. "26" + "1234" = "261234" ($$$2+6+1 = 9 = 2 + 3 + 4$$$).

    Define count(len, sum) as the number of snippets in the input that have length len and the sum of the digits is sum. It's easy to precalculate and store these in an array count[6][46] or something (since length is limited to 5, and therefore sum of digits is limited to 45).

    Now for each snippet, consider how many of each of the three cases you can generate.

    For case 1, a snippet s with length l and sum s can appear on the left side exactly count(l, s) times (i.e., it can be paired with each snippet [including itself!] that has the same length and digit sum). That's the easy case. (You might ask: what about when s appears on the right side? Ignore that case to avoid double counting.)

    For case 2, a prefix of s of length k will be the first half, and the remaining digits will be part of the right half. So for example, if we have s="1234567" which has length 7, we can construct tickets "123456|7abcde" or "12345|67abc" or "1234|567a" (where '|' marks the middle). Consider the first one: "123456|7abcde". Here, the left side has sum 1+2+3+4+5+6 = 15, so the right side must have the same sum. The right side already has a 7, so the remaining 5 digits (*abcde*) must have sum 15-7=8. So the number of tickets of the form "123456|7abcde" is count(12-7, 1+2+3+4+5+6-7). Similarly, the number of tickets of the form "12345|67abc" = count(10-7, (1+2+3+4+5)-(6+7)) You can generalize this to something like: $$$count(|s| - k, \sum s_{1..k} - \sum s_{k..|s|})$$$ for all $$$|s|/2 < k < |s|$$$.

    Case 3 is exactly the same as case 2 but using suffixes instead of prefixes.

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      7 months ago, # ^ |
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      My code is here plz . this is my submission.

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        7 months ago, # ^ |
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        Your code is not in a proper format. Better you add submission link instead of whole code.

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          7 months ago, # ^ |
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          I have provided the link . My code is here plz this ( it is the link)

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            7 months ago, # ^ |
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            a if condn was missing , now its solved . i aim to become expert by the end of this year , and do everything possible to acheive it

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this contest proved how bad i am at implementation.

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Can someone provide test cases for problem C? I'm struggling to come up with good ones to test my code. Thank you.1895C - Порванный счастливый билет 231207578

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7 months ago, # |
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Was anyone able to pass a randomized solution for problem D?

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ok, this is my first time hacking submissions in contests

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how to solve C?i have no idea? need help

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    7 months ago, # ^ |
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    For each number, calculate the length,the length of the first half and the sum of digits in the first half of the prefix it needs as a suffix,and put them in a three-dimensional bucket. Then use all the numbers to match these as prefix. 231178050

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How to solve E?__

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    7 months ago, # ^ |
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    The problem says that once a card is beaten, the player draws it back, so it is always optimal to play the best card $$$b_j$$$ against a card $$$a_i$$$ such that attack of $$$b_j$$$ > defense of $$$a_i$$$, and the defense of $$$b_j$$$ is maximum among such possible cards. This means we can make directed edges among the cards of Alice and Bob.

    Those cards for which there are no possible opponent cards which beats them, are considered to end the game. Now, as we have a directed graph, we can easily use dfs to find out which the result for every card $$$a_i$$$. Note that, for every component in the directed graph, there would be at most one cycle in that component.

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How to solve F?

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In the question D of this contest, when can ai be equals to 2*n.

Is it even possible, If yes then the case??

As max value of A^B is A+B but not for A==B. Please help

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    7 months ago, # ^ |
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    it's impossible but he promised that it's always possible to construct at least one valid array b from the given sequence a.

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      7 months ago, # ^ |
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      I figured out that we only need to find b1 as all other values can be derived from b1 since b2 = a1 ^ b1, b3 = a1 ^ a2 ^ b1 ... bN = a1 ^ a2 ^ ... ^ aN-1 ^ b1. But I can't find fast way to determine if a b1 if valid how did you do that or if you have any other idea?

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        7 months ago, # ^ |
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        You can just enumerate it, and try to check the answer in $$$O(\log n)$$$.

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7 months ago, # |
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It says rated for people under 2100 but no changes so far in rating

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7 months ago, # |
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why am i still unrated for this contest?

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    7 months ago, # ^ |
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    now it is showing that i haven't even given this contest . What is happening?

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7 months ago, # |
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What happened on problem E? Why are there many judgement failed?

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What's wrong with problem E?

Everyone got Judgement failed in system test.

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    7 months ago, # ^ |
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    i think the standard solution of problem e may be wrong

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7 months ago, # |
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now it is showing the contest but no change in rating.

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    7 months ago, # ^ |
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    That is common. All of these kind of contest(div3, div4 and edu) have an open hack. And the rating change will be much slower.

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      7 months ago, # ^ |
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      oh thanks . But will it show unrated until the rating is changed or does it show unrated if the rating won't change at all.

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        7 months ago, # ^ |
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        It show unrated until the rating is changed.

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problem can easily be solved by https://www.geeksforgeeks.org/find-maximum-xor-given-integer-stream-integers/ and i read some comments saying that it was like another problems i think thats not ok that problems repeated

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    7 months ago, # ^ |
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    I think D is not repeated because the main focus of this question is to transform it into "find the max after xor given integer". It is easy to solve the latter lol.

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    7 months ago, # ^ |
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    The reduction to this basic 01trie trick requires some non-obvious steps, such as taking the prefix sum, then proving that the prefix sum must be unique, and finally reducing it to the condition of max XOR-sum $$$\leq n-1$$$. And also that problem on leetcode only involves the case where $$$n$$$ is odd, which is trivial and imo does not provide additional insight to the even case.

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Why are they not judging E? Did someone TLE/RE Hacked the main solution or smth?

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    7 months ago, # ^ |
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    It was a hack with a java21 generator, but the judging system wasn't re-configured to support them. I already fixed it.

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Is the educational round unrated? The rating is still not updated..... It's my first educational round.... Someone please tell me

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7 months ago, # |
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Isn't it a bit late for not putting Editorial?

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7 months ago, # |
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Why not release full 5-hour version of a regional contest? What's the point in spoiling half of the problems?

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    7 months ago, # ^ |
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    I think 5h for a rated contest is too much because of time zone difference and usual life stuff that might take over and not let many fully concentrate on the contest for so long

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7 months ago, # |
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For F,why can (k — x + 1) be more than n.The situation doesn't follow condition 1.

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6 months ago, # |
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MikeMirzayanov and awoo, User ALL_LOWERCASED, Sir-Ahmed-Imran and I got the solution of the problem D skipped because of similarities. But the code of the Trie which was common was taken from this website, the code which we used is at the bottom of the page.The code was available before the contest.